The Notes: `int_0^0.6 x^2 /sqrt(9 - 25x^2) dx` Evaluate the integral

Thursday, March 26, 2009

$\displaystyle{\int_{{0}}^{{0.6}}}\frac{{{x}}^{{2}}}{\sqrt{{{9}-{25}{{x}}^{{2}}}}}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle{\int_{{0}}^{{0.6}}}\frac{{{x}}^{{2}}}{\sqrt{{{9}-{25}{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

Let's first evaluate the indefinite integral by applying the integral substitution,

Let $\displaystyle{x}=\frac{{3}}{{5}}{\sin{{\left({u}\right)}}}$

$\displaystyle\Rightarrow{\left.{d}{x}\right.}=\frac{{3}}{{5}}{\cos{{\left({u}\right)}}}{d}{u}$

Plug in the above in the integral,

$\displaystyle\int\frac{{{x}}^{{2}}}{\sqrt{{{9}-{25}{{x}}^{{2}}}}}{\left.{d}{x}\right.}=\int\frac{{{\left(\frac{{3}}{{5}}{\sin{{\left({u}\right)}}}\right)}}^{{2}}}{{\sqrt{{{9}-{25}{{\left(\frac{{3}}{{5}}{\sin{{\left({u}\right)}}}\right)}}^{{2}}}}}}{\left(\frac{{3}}{{5}}{\cos{{\left({u}\right)}}}\right)}{d}{u}$

$\displaystyle=\int\frac{{{27}{{\sin}}^{{2}}{\left({u}\right)}{\cos{{\left({u}\right)}}}}}{{{125}\sqrt{{{9}-{9}{{\sin}}^{{2}}{\left({u}\right)}}}}}{d}{u}$

$\displaystyle=\int\frac{{{27}{{\sin}}^{{2}}{\left({u}\right)}{\cos{{\left({u}\right)}}}}}{{{125}\sqrt{{{9}{\left({1}-{{\sin}}^{{2}}{\left({u}\right)}\right)}}}}}{d}{u}$

use the identity: $\displaystyle{1}-{{\sin}}^{{2}}{\left({x}\right)}={{\cos}}^{{2}}{\left({x}\right)}$ $\displaystyle<{b}{r}>$

$\displaystyle=\int\frac{{{27}{{\sin}}^{{2}}{\left({u}\right)}{\cos{{\left({u}\right)}}}}}{{{125}\cdot{3}\sqrt{{{{\cos}}^{{2}}{\left({u}\right)}}}}}{d}{u}$

$\displaystyle=\int\frac{{{9}{{\sin}}^{{2}}{\left({u}\right)}{\cos{{\left({u}\right)}}}}}{{{125}{\cos{{\left({u}\right)}}}}}{d}{u}$

$\displaystyle=\frac{{9}}{{125}}\int{{\sin}}^{{2}}{\left({u}\right)}{d}{u}$

Now use the identity: $\displaystyle{{\sin}}^{{2}}{\left({x}\right)}=\frac{{{1}-{\cos{{\left({2}{x}\right)}}}}}{{2}}$

$\displaystyle=\frac{{9}}{{125}}\int\frac{{{1}-{\cos{{\left({2}{u}\right)}}}}}{{2}}{d}{u}$

$\displaystyle=\frac{{9}}{{250}}\int{\left({1}-{\cos{{\left({2}{u}\right)}}}\right)}{d}{u}$

$\displaystyle=\frac{{9}}{{250}}{\left(\int{1}{d}{u}-\int{\cos{{\left({2}{u}\right)}}}{d}{u}\right)}$

$\displaystyle=\frac{{9}}{{250}}{\left({u}-\frac{{\sin{{\left({2}{u}\right)}}}}{{2}}\right)}$

We have taken $\displaystyle{x}=\frac{{3}}{{5}}{\sin{{\left({u}\right)}}}$

$\displaystyle\Rightarrow{u}={\arcsin{{\left(\frac{{5}}{{3}}{x}\right)}}}$

Substitute back u and add a constant C to the solution,

$\displaystyle=\frac{{9}}{{250}}{\left({\arcsin{{\left(\frac{{5}}{{3}}{x}\right)}}}-\frac{{1}}{{2}}{\sin{{\left({2}{\arcsin{{\left(\frac{{5}}{{3}}{x}\right)}}}\right)}}}\right)}+{C}$

Now let's evaluate the definite integral,

$\displaystyle{\int_{{0}}^{{0.6}}}\frac{{{x}}^{{2}}}{\sqrt{{{9}-{25}{{x}}^{{2}}}}}{\left.{d}{x}\right.}={{\left[\frac{{9}}{{250}}{\left({\arcsin{{\left(\frac{{5}}{{3}}{x}\right)}}}-\frac{{1}}{{2}}{\sin{{\left({2}{\arcsin{{\left(\frac{{5}}{{3}}{x}\right)}}}\right)}}}\right)}\right]}_{{0}}^{{0.6}}}$

$\displaystyle={\left[\frac{{9}}{{250}}{\left({\arcsin{{\left(\frac{{5}}{{3}}\cdot{0.6}\right)}}}-\frac{{1}}{{2}}{\sin{{\left({2}{\arcsin{{\left(\frac{{5}}{{3}}\cdot{0.6}\right)}}}\right)}}}\right)}\right]}-{\left[\frac{{9}}{{250}}{\left({\arcsin{{\left(\frac{{5}}{{3}}\cdot{0}\right)}}}-\frac{{1}}{{2}}{\sin{{\left({2}{\arcsin{{\left(\frac{{5}}{{3}}\cdot{0}\right)}}}\right)}}}\right)}\right]}$

$\displaystyle={\left[\frac{{9}}{{250}}{\left({\arcsin{{\left({1}\right)}}}-\frac{{1}}{{2}}{\sin{{\left({2}{\arcsin{{\left({1}\right)}}}\right)}}}\right)}\right]}-{\left[\frac{{9}}{{250}}{\left({\arcsin{{\left({0}\right)}}}-\frac{{1}}{{2}}{\sin{{\left({2}{\arcsin{{\left({0}\right)}}}\right)}}}\right)}\right]}$

$\displaystyle={\left[\frac{{9}}{{250}}{\left(\frac{\pi}{{2}}-\frac{{1}}{{2}}{\sin{{\left({2}\cdot\frac{\pi}{{2}}\right)}}}\right)}\right]}-{\left[{0}\right]}$